\(\int \frac {A+B x}{x^4 (a+b x)^3} \, dx\) [170]

Optimal result

Integrand size = 16, antiderivative size = 140 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=-\frac {A}{3 a^3 x^3}+\frac {3 A b-a B}{2 a^4 x^2}-\frac {3 b (2 A b-a B)}{a^5 x}-\frac {b^2 (A b-a B)}{2 a^4 (a+b x)^2}-\frac {b^2 (4 A b-3 a B)}{a^5 (a+b x)}-\frac {2 b^2 (5 A b-3 a B) \log (x)}{a^6}+\frac {2 b^2 (5 A b-3 a B) \log (a+b x)}{a^6} \] Output:

-1/3*A/a^3/x^3+1/2*(3*A*b-B*a)/a^4/x^2-3*b*(2*A*b-B*a)/a^5/x-1/2*b^2*(A*b- 
B*a)/a^4/(b*x+a)^2-b^2*(4*A*b-3*B*a)/a^5/(b*x+a)-2*b^2*(5*A*b-3*B*a)*ln(x) 
/a^6+2*b^2*(5*A*b-3*B*a)*ln(b*x+a)/a^6
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=\frac {\frac {a \left (-60 A b^4 x^4+18 a b^3 x^3 (-5 A+2 B x)-a^4 (2 A+3 B x)+a^3 b x (5 A+12 B x)+2 a^2 b^2 x^2 (-10 A+27 B x)\right )}{x^3 (a+b x)^2}+12 b^2 (-5 A b+3 a B) \log (x)+12 b^2 (5 A b-3 a B) \log (a+b x)}{6 a^6} \] Input:

Integrate[(A + B*x)/(x^4*(a + b*x)^3),x]
 

Output:

((a*(-60*A*b^4*x^4 + 18*a*b^3*x^3*(-5*A + 2*B*x) - a^4*(2*A + 3*B*x) + a^3 
*b*x*(5*A + 12*B*x) + 2*a^2*b^2*x^2*(-10*A + 27*B*x)))/(x^3*(a + b*x)^2) + 
 12*b^2*(-5*A*b + 3*a*B)*Log[x] + 12*b^2*(5*A*b - 3*a*B)*Log[a + b*x])/(6* 
a^6)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {86, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B x}{x^4 (a+b x)^3} \, dx\)

\(\Big \downarrow \) 86

\(\displaystyle \int \left (-\frac {2 b^3 (3 a B-5 A b)}{a^6 (a+b x)}+\frac {2 b^2 (3 a B-5 A b)}{a^6 x}-\frac {b^3 (3 a B-4 A b)}{a^5 (a+b x)^2}-\frac {3 b (a B-2 A b)}{a^5 x^2}-\frac {b^3 (a B-A b)}{a^4 (a+b x)^3}+\frac {a B-3 A b}{a^4 x^3}+\frac {A}{a^3 x^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b^2 \log (x) (5 A b-3 a B)}{a^6}+\frac {2 b^2 (5 A b-3 a B) \log (a+b x)}{a^6}-\frac {b^2 (4 A b-3 a B)}{a^5 (a+b x)}-\frac {3 b (2 A b-a B)}{a^5 x}-\frac {b^2 (A b-a B)}{2 a^4 (a+b x)^2}+\frac {3 A b-a B}{2 a^4 x^2}-\frac {A}{3 a^3 x^3}\)

Input:

Int[(A + B*x)/(x^4*(a + b*x)^3),x]
 

Output:

-1/3*A/(a^3*x^3) + (3*A*b - a*B)/(2*a^4*x^2) - (3*b*(2*A*b - a*B))/(a^5*x) 
 - (b^2*(A*b - a*B))/(2*a^4*(a + b*x)^2) - (b^2*(4*A*b - 3*a*B))/(a^5*(a + 
 b*x)) - (2*b^2*(5*A*b - 3*a*B)*Log[x])/a^6 + (2*b^2*(5*A*b - 3*a*B)*Log[a 
 + b*x])/a^6
 

Defintions of rubi rules used

rule 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {A}{3 a^{3} x^{3}}-\frac {-3 A b +B a}{2 x^{2} a^{4}}-\frac {3 b \left (2 A b -B a \right )}{a^{5} x}-\frac {2 b^{2} \left (5 A b -3 B a \right ) \ln \left (x \right )}{a^{6}}-\frac {b^{2} \left (4 A b -3 B a \right )}{a^{5} \left (b x +a \right )}-\frac {b^{2} \left (A b -B a \right )}{2 a^{4} \left (b x +a \right )^{2}}+\frac {2 b^{2} \left (5 A b -3 B a \right ) \ln \left (b x +a \right )}{a^{6}}\)	\(134\)
norman	\(\frac {-\frac {A}{3 a}+\frac {\left (5 A b -3 B a \right ) x}{6 a^{2}}-\frac {2 b \left (5 A b -3 B a \right ) x^{2}}{3 a^{3}}-\frac {2 \left (5 b^{5} A -3 a \,b^{4} B \right ) x^{4}}{a^{5} b}-\frac {\left (15 b^{5} A -9 a \,b^{4} B \right ) x^{3}}{a^{4} b^{2}}}{x^{3} \left (b x +a \right )^{2}}-\frac {2 b^{2} \left (5 A b -3 B a \right ) \ln \left (x \right )}{a^{6}}+\frac {2 b^{2} \left (5 A b -3 B a \right ) \ln \left (b x +a \right )}{a^{6}}\)	\(145\)
risch	\(\frac {-\frac {2 b^{3} \left (5 A b -3 B a \right ) x^{4}}{a^{5}}-\frac {3 b^{2} \left (5 A b -3 B a \right ) x^{3}}{a^{4}}-\frac {2 b \left (5 A b -3 B a \right ) x^{2}}{3 a^{3}}+\frac {\left (5 A b -3 B a \right ) x}{6 a^{2}}-\frac {A}{3 a}}{x^{3} \left (b x +a \right )^{2}}+\frac {10 b^{3} \ln \left (-b x -a \right ) A}{a^{6}}-\frac {6 b^{2} \ln \left (-b x -a \right ) B}{a^{5}}-\frac {10 b^{3} \ln \left (x \right ) A}{a^{6}}+\frac {6 b^{2} \ln \left (x \right ) B}{a^{5}}\)	\(151\)
parallelrisch	\(-\frac {60 A \ln \left (x \right ) x^{5} b^{7}-60 A \ln \left (b x +a \right ) x^{5} b^{7}-36 B \ln \left (x \right ) x^{5} a \,b^{6}+36 B \ln \left (b x +a \right ) x^{5} a \,b^{6}+120 A \ln \left (x \right ) x^{4} a \,b^{6}-120 A \ln \left (b x +a \right ) x^{4} a \,b^{6}-72 B \ln \left (x \right ) x^{4} a^{2} b^{5}+72 B \ln \left (b x +a \right ) x^{4} a^{2} b^{5}+60 A \ln \left (x \right ) x^{3} a^{2} b^{5}-60 A \ln \left (b x +a \right ) x^{3} a^{2} b^{5}+60 A \,x^{4} a \,b^{6}-36 B \ln \left (x \right ) x^{3} a^{3} b^{4}+36 B \ln \left (b x +a \right ) x^{3} a^{3} b^{4}-36 B \,x^{4} a^{2} b^{5}+90 A \,x^{3} a^{2} b^{5}-54 B \,x^{3} a^{3} b^{4}+20 A \,x^{2} a^{3} b^{4}-12 B \,x^{2} a^{4} b^{3}-5 A x \,a^{4} b^{3}+3 B x \,a^{5} b^{2}+2 A \,a^{5} b^{2}}{6 a^{6} b^{2} x^{3} \left (b x +a \right )^{2}}\)	\(297\)

Input:

int((B*x+A)/x^4/(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/3*A/a^3/x^3-1/2*(-3*A*b+B*a)/x^2/a^4-3*b*(2*A*b-B*a)/a^5/x-2*b^2*(5*A*b 
-3*B*a)*ln(x)/a^6-b^2*(4*A*b-3*B*a)/a^5/(b*x+a)-1/2*b^2*(A*b-B*a)/a^4/(b*x 
+a)^2+2*b^2*(5*A*b-3*B*a)*ln(b*x+a)/a^6
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.87 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=-\frac {2 \, A a^{5} - 12 \, {\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} - 18 \, {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} - 4 \, {\left (3 \, B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (3 \, B a^{5} - 5 \, A a^{4} b\right )} x + 12 \, {\left ({\left (3 \, B a b^{4} - 5 \, A b^{5}\right )} x^{5} + 2 \, {\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3}\right )} \log \left (b x + a\right ) - 12 \, {\left ({\left (3 \, B a b^{4} - 5 \, A b^{5}\right )} x^{5} + 2 \, {\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}} \] Input:

integrate((B*x+A)/x^4/(b*x+a)^3,x, algorithm="fricas")
 

Output:

-1/6*(2*A*a^5 - 12*(3*B*a^2*b^3 - 5*A*a*b^4)*x^4 - 18*(3*B*a^3*b^2 - 5*A*a 
^2*b^3)*x^3 - 4*(3*B*a^4*b - 5*A*a^3*b^2)*x^2 + (3*B*a^5 - 5*A*a^4*b)*x + 
12*((3*B*a*b^4 - 5*A*b^5)*x^5 + 2*(3*B*a^2*b^3 - 5*A*a*b^4)*x^4 + (3*B*a^3 
*b^2 - 5*A*a^2*b^3)*x^3)*log(b*x + a) - 12*((3*B*a*b^4 - 5*A*b^5)*x^5 + 2* 
(3*B*a^2*b^3 - 5*A*a*b^4)*x^4 + (3*B*a^3*b^2 - 5*A*a^2*b^3)*x^3)*log(x))/( 
a^6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3)
 

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.87 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=\frac {- 2 A a^{4} + x^{4} \left (- 60 A b^{4} + 36 B a b^{3}\right ) + x^{3} \left (- 90 A a b^{3} + 54 B a^{2} b^{2}\right ) + x^{2} \left (- 20 A a^{2} b^{2} + 12 B a^{3} b\right ) + x \left (5 A a^{3} b - 3 B a^{4}\right )}{6 a^{7} x^{3} + 12 a^{6} b x^{4} + 6 a^{5} b^{2} x^{5}} + \frac {2 b^{2} \left (- 5 A b + 3 B a\right ) \log {\left (x + \frac {- 10 A a b^{3} + 6 B a^{2} b^{2} - 2 a b^{2} \left (- 5 A b + 3 B a\right )}{- 20 A b^{4} + 12 B a b^{3}} \right )}}{a^{6}} - \frac {2 b^{2} \left (- 5 A b + 3 B a\right ) \log {\left (x + \frac {- 10 A a b^{3} + 6 B a^{2} b^{2} + 2 a b^{2} \left (- 5 A b + 3 B a\right )}{- 20 A b^{4} + 12 B a b^{3}} \right )}}{a^{6}} \] Input:

integrate((B*x+A)/x**4/(b*x+a)**3,x)
                                                                                    
                                                                                    
 

Output:

(-2*A*a**4 + x**4*(-60*A*b**4 + 36*B*a*b**3) + x**3*(-90*A*a*b**3 + 54*B*a 
**2*b**2) + x**2*(-20*A*a**2*b**2 + 12*B*a**3*b) + x*(5*A*a**3*b - 3*B*a** 
4))/(6*a**7*x**3 + 12*a**6*b*x**4 + 6*a**5*b**2*x**5) + 2*b**2*(-5*A*b + 3 
*B*a)*log(x + (-10*A*a*b**3 + 6*B*a**2*b**2 - 2*a*b**2*(-5*A*b + 3*B*a))/( 
-20*A*b**4 + 12*B*a*b**3))/a**6 - 2*b**2*(-5*A*b + 3*B*a)*log(x + (-10*A*a 
*b**3 + 6*B*a**2*b**2 + 2*a*b**2*(-5*A*b + 3*B*a))/(-20*A*b**4 + 12*B*a*b* 
*3))/a**6
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=-\frac {2 \, A a^{4} - 12 \, {\left (3 \, B a b^{3} - 5 \, A b^{4}\right )} x^{4} - 18 \, {\left (3 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{3} - 4 \, {\left (3 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + {\left (3 \, B a^{4} - 5 \, A a^{3} b\right )} x}{6 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} - \frac {2 \, {\left (3 \, B a b^{2} - 5 \, A b^{3}\right )} \log \left (b x + a\right )}{a^{6}} + \frac {2 \, {\left (3 \, B a b^{2} - 5 \, A b^{3}\right )} \log \left (x\right )}{a^{6}} \] Input:

integrate((B*x+A)/x^4/(b*x+a)^3,x, algorithm="maxima")
 

Output:

-1/6*(2*A*a^4 - 12*(3*B*a*b^3 - 5*A*b^4)*x^4 - 18*(3*B*a^2*b^2 - 5*A*a*b^3 
)*x^3 - 4*(3*B*a^3*b - 5*A*a^2*b^2)*x^2 + (3*B*a^4 - 5*A*a^3*b)*x)/(a^5*b^ 
2*x^5 + 2*a^6*b*x^4 + a^7*x^3) - 2*(3*B*a*b^2 - 5*A*b^3)*log(b*x + a)/a^6 
+ 2*(3*B*a*b^2 - 5*A*b^3)*log(x)/a^6
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=\frac {2 \, {\left (3 \, B a b^{2} - 5 \, A b^{3}\right )} \log \left ({\left | x \right |}\right )}{a^{6}} - \frac {2 \, {\left (3 \, B a b^{3} - 5 \, A b^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{6} b} - \frac {2 \, A a^{5} - 12 \, {\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} - 18 \, {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} - 4 \, {\left (3 \, B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (3 \, B a^{5} - 5 \, A a^{4} b\right )} x}{6 \, {\left (b x + a\right )}^{2} a^{6} x^{3}} \] Input:

integrate((B*x+A)/x^4/(b*x+a)^3,x, algorithm="giac")
 

Output:

2*(3*B*a*b^2 - 5*A*b^3)*log(abs(x))/a^6 - 2*(3*B*a*b^3 - 5*A*b^4)*log(abs( 
b*x + a))/(a^6*b) - 1/6*(2*A*a^5 - 12*(3*B*a^2*b^3 - 5*A*a*b^4)*x^4 - 18*( 
3*B*a^3*b^2 - 5*A*a^2*b^3)*x^3 - 4*(3*B*a^4*b - 5*A*a^3*b^2)*x^2 + (3*B*a^ 
5 - 5*A*a^4*b)*x)/((b*x + a)^2*a^6*x^3)
 

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=\frac {4\,b^2\,\mathrm {atanh}\left (\frac {2\,b^2\,\left (5\,A\,b-3\,B\,a\right )\,\left (a+2\,b\,x\right )}{a\,\left (10\,A\,b^3-6\,B\,a\,b^2\right )}\right )\,\left (5\,A\,b-3\,B\,a\right )}{a^6}-\frac {\frac {A}{3\,a}-\frac {x\,\left (5\,A\,b-3\,B\,a\right )}{6\,a^2}+\frac {3\,b^2\,x^3\,\left (5\,A\,b-3\,B\,a\right )}{a^4}+\frac {2\,b^3\,x^4\,\left (5\,A\,b-3\,B\,a\right )}{a^5}+\frac {2\,b\,x^2\,\left (5\,A\,b-3\,B\,a\right )}{3\,a^3}}{a^2\,x^3+2\,a\,b\,x^4+b^2\,x^5} \] Input:

int((A + B*x)/(x^4*(a + b*x)^3),x)
 

Output:

(4*b^2*atanh((2*b^2*(5*A*b - 3*B*a)*(a + 2*b*x))/(a*(10*A*b^3 - 6*B*a*b^2) 
))*(5*A*b - 3*B*a))/a^6 - (A/(3*a) - (x*(5*A*b - 3*B*a))/(6*a^2) + (3*b^2* 
x^3*(5*A*b - 3*B*a))/a^4 + (2*b^3*x^4*(5*A*b - 3*B*a))/a^5 + (2*b*x^2*(5*A 
*b - 3*B*a))/(3*a^3))/(a^2*x^3 + b^2*x^5 + 2*a*b*x^4)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.69 \[ \int \frac {A+B x}{x^4 (a+b x)^3} \, dx=\frac {12 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}+12 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}-12 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{3}-12 \,\mathrm {log}\left (x \right ) b^{4} x^{4}-a^{4}+2 a^{3} b x -6 a^{2} b^{2} x^{2}+12 b^{4} x^{4}}{3 a^{5} x^{3} \left (b x +a \right )} \] Input:

int((B*x+A)/x^4/(b*x+a)^3,x)
 

Output:

(12*log(a + b*x)*a*b**3*x**3 + 12*log(a + b*x)*b**4*x**4 - 12*log(x)*a*b** 
3*x**3 - 12*log(x)*b**4*x**4 - a**4 + 2*a**3*b*x - 6*a**2*b**2*x**2 + 12*b 
**4*x**4)/(3*a**5*x**3*(a + b*x))